TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 170, August 1972
STONE'S TOPOLOGY FOR PSEUDOCOMPLEMENTED AND BICOMPLEMENTED LATTICES^)
BY P. V. VENKATANARASIMHAN
ABSTRACT. In an earlier paper the author has proved the existence of prime ideals and prime dual ideals in a pseudocomplemented lattice (not necessarily distributive). The present paper is devoted to a study of Stone's topology on the set of prime dual ideals of a pseudocomplemented and a bicomplemented lattice. If L is the quotient lattice arising out of the congruence relation defined by a = b ^^ a* = b* in a pseudocomplemented lattice L, it is proved that Stone's space of prime dual ideals of L is homeomorphic to the subspace of maximal dual ideals of L.
Stone [9] has introduced a topology for the set of all prime ideals of a distrib- utive lattice. Balachandran [3] has made an extensive study of Stone's topology of the distributive lattice and has obtained results supplementing those of Stone. The purpose of this paper is to extend some of the results of Stone and Balachan- dran to pseudocomplemented and bicomplemented lattices. (A lattice closed for pseudocomplements as well as quasi complements is called a bicomplemented lat- tice.) In the first section we collect some known results which are used in subse- quent sections. §§2 and 3 deal with Stone's topology on the set of prime dual ideals of a pseudocomplemented lattice and a bicomplemented lattice respective- ly. The concluding section is devoted to a study of ideals and dual ideals of the quotient lattice of a pseudocomplemented lattice with respect to a special congru- ence relation. It is proved that Stone's space of prime dual ideals of the quotient lattice is homeomorphic to the subspace of maximal dual ideals of the given lat- tice. The author is thankful to the referee for various suggestions which helped to improve the earlier version of the paper. In particular, he is indebted to the refer- ee for pointing out that the earlier versions of some of the results were not valid.
Received by the editors June 3, 1970 and, in revised form, January 11, 1971 and Oc- tober 15, 1971. AMS 1969 subject classifications. Primary 0635; Secondary 0650. Key words and phrases. Pseudocomplemented lattice, bicomplemented lattice, distri- butive lattice, Boolean algebra, normal element, simple element, prime ideal, prime dual ideal, quotient lattice, Stone topology. (1) This paper formed a part of the author's Ph.D. Thesis (University of Madras) pre- pared under the guidance of Professors V. S. Krishnan and V. K. Balachandran. Copyright © 1972, American Mathematical Society 57
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1. Preliminaries. For the lattice-theoretic and topological concepts which have now become commonplace the reader is referred to [4], [6] and [lO]. We shall recall some concepts which are not quite well known. Let a be an element of a semilattice S with 0. Then an element a* of S is called a pseudocomplement of a if (i) aa* = 0 and (ii) ab = 0 =^ b < a*. An ele- ment ä of a dual semilattice S with 1 is called a quasicomplement of an element a of S if (i) a + a = 1 and (ii) a + ¿>= 1 =^ è > a. A semilattice in which every element has a pseudocomplement is said to be pseudocomplemented. Similarly we define a quasicomplemented dual semilattice. An element a of a pseudocomple- mented semilattice is said to be normal (dense) if a = a** (a* = 0). A subset A of a partially ordered set P is called a semi-ideal ii a £ A, b < a => b £ A. A semi-ideal A of P is called an ideal if the lattice sum of any finite number of elements of A, whenever it exists, belongs to A. Dual semi-ideal and dual ideal ate defined in a dual fashion. Obviously, in a lattice the above definitions coin- cide with the usual definitions. An element a of a pseudocomplemented lattice is said to be simple if a + a* = 1. A pseudocomplemented lattice in which every normal element is simple is called an S-lattice. A distributive S-lattice is called a Stone lattice. The results summarized in the form of Theorem I below are due to Frink [5].
Theorem I. In any pseudocomplemented semilattice S, the following results hold: (i) a < a**, (ii) a*** = a*, (iii) a <; b =£» a* ~¿ b*, (iv) (ab)* = (a**b**)*, (v) (ab)** = a**b**, (vi) S has the greatest element 1 and 0* = 1, (vii) a is nor- mal if and only if a = b* for some b, (viii) the set N of normal elements of S forms a Boolean algebra; the lattice sum of any two elements a, b of N is (a*b*)* and their lattice product is the same as that in L, (ix) if L is a lattice, (a + b)*= a*b* and a* + b* < (ab)*.
Let X be a topological space. A point p of X is said to be Tx (anti-Tx) if q i closure of p (p / closure of q) for any point q of X other than p. X is said to be Il0 if every non-null open subset of X contains a non-null closed subset. Let A and B be two disjoint subsets of X. Then we say A is weakly separable from B if there exists an open subset of X containing A and disjoint with B. It is easily seen that A is weakly separable from B if and only if AriclB =
Lemma I [13]. Any proper dual ideal of a partially ordered set with 0 is con- tained in a maximal dual ideal.
Lemma II [14]. Any maximal dual ideal of a pseudocomplemented lattice is prime.
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Lemma III [13]. The set D of all dense elements of a partially ordered set P with 0,1 forms a dual ideal and D is the product of all the maximal dual ideals of P.
Lemma IV [14]. Any prime ideal of a pseudocomplemented semilattice con- tains a minimal prime ideal.
Lemma V [ll]. // in a modular lattice, the lattice sum and lattice product of two ideals are principal, then the ideals themselves are principal.
Lemma VI [14]. A prime ideal of a pseudocomplemented semilattice S is min- imal prime if and only if it contains precisely one of x, x* for every x in S.
For convenience we shall prove the 'only if part of Lemma VI here. First we observe that any maximal dual ideal M of S contains precisely one of x, x* for every x in S. (Since xx* = 0, it is clear that M contains at most one of x, x*. If x ^ M, then M V [x) = 5. Hence yx = 0 for some y in M; y £ x* and so x* £ M,) Now we shall prove that if A is a minimal prime ideal of S, then the set complement cA oí A is a maximal dual ideal. Clearly cA is a dual ideal. By Lemma I, cA is contained in a maximal dual ideal M of S. AD cM and clearly cM is a semi-ideal. Let x x2, • • • , x £ cM and suppose x, + x2 + • • • + x exists. Since M is a maximal dual ideal, x* x*, • • • , x* £ M and so x°Çx* • • • x* £ M. That is (x, + x, + • • •+ x )* £ M. Hence x, + x, + • • • + x £ cM. Thus cM 12 n 12 n is an ideal. It is easily seen that cM is prime. By the minimality of A, it fol- lows that A = cM. Hence cA = M, so that cA is a maximal dual ideal. From the above it follows that every minimal prime ideal of S contains precisely one of x, x* for every x in S.
Lemma VII [ll]. // in a pseudocomplemented modular lattice L, D = [l), then L is a Boolean algebra.
Lemma VIII. The first three of the following statements concerning a pseudo- complemented lattice L are equivalent and each of these is implied by the fourth. (i) Every prime ideal is minimal prime. (ii) Every prime dual ideal is minimal prime. (iii) Every prime dual ideal is maximal. (iv) £>= [1).
Proof. Suppose A is a prime dual ideal which is not minimal prime. Then there exists a prime dual ideal B such that B Ç A; cA, cB ate prime ideals and cA r cB (cA, cB ate the set complements of A, B respectively). It follows that cB is not minimal prime. Thus, (i) =^ (ii). Let C be a prime dual ideal which is not maximal. Then by Lemma I, there
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exists a maximal dual ideal M such that M D C. By Lemma II, M is prime. Thus M is a prime dual ideal which is not minimal prime. Hence (ii) ^» (iii). Let A be a prime ideal which is not minimal prime. Then by Lemma IV, there is a minimal prime ideal B such that B Ç A. Clearly cA and cB ate proper prime dual ideals and cB 3 cA. Thus cA is a prime dual ideal which is not maxi- mal. Hence (iii) =#►(i). Suppose (iv) holds. Then for every a £ L, a + a* = 1 and so every prime ideal contains precisely one of a, a*. Hence by Lemma VI, every prime ideal is minimal prime. Thus (iv) =^ (i).
2. Stone's topology of the pseudocomplemented lattice. In this section we extend some results of Stone [9] and Balachandran [3] on Stone's topology of the distributive lattice to pseudocomplemented lattices. Throughout this section L denotes a pseudocomplemented lattice and S) the lattice of dual ideals of L. We denote by J the set of all prime dual ideals of L. For any dual ideal A of L, F(A) denotes the set of all prime dual ideals of L containing A and F' (A) the set complement of F (A) in J. The lattice sum and lattice product of two elements a, b of L are denoted by a + b and ab respectively. The pseudocomplement of a is denoted by a*. We shall denote the lattice sums in the lattice of all ideals as well as the lattice of all dual ideals of L by V. The lattice products in these lattices coincide with the corresponding set intersections. Set inclusion, set union and set intersection are denoted by C, U and Pi respectively. Theorems 1, 2 and 3 below, enunciated without proofs, are analogues of the corresponding results in §6 of [12] and the proofs of these are similar to those of the corresponding results of [12] with obvious modifications.
Theorem 1. (i) F(\/i£lAi)=r\i£lF(Ai), (ii) F(A x n A2 O . . • O Aj = F (A x) UF(A2)U...U F(AJ, (iii) F(L) =
Corollary, (i) F'( V,-,, A¿) = Ui£,F'(A), (ii) F'(A x n A2n .. .n Aj = F'(Ax) n f'(a2)n...nF,{An), (iii) F'(D = f, (iv) F'([l)) = 0.
Theorem 1 shows that F defines a closure operation in J thereby giving rise to a topology T on J.
Theorem 2. The lattice of all open (closed) sets of (?, T) is homomorphic (dually homomorphic) to 2) and the mapping A —» F'(A) (A —» F(A)) takes arbi- trary lattice sums to corresponding set unions (set intersections).
Theorem 3. // X is any subset of 9, clX = F(XQ), XQ being the product of all the members of X.
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Theorem 4. (3\ T) is TQ.
Theorem 4 follows immediately from Theorem 3.
Theorem 5. {j, T) is compact. Proof. Let 9= U¿e/F'U¿). Then by the corollary under Theorem 1, F'{L) = F ( V¿e/A¿). Therefore Wie¡A¿ = L (for, otherwise by Lemmas I and II, there would exist a prime dual ideal containing V , A . which is a contradiction to the fact that F \v-ejA) = j). Hence there exist a finite number of elements a¿., «¡,, •••,«,• (a¿.eA¿.) suchthat 0 = ß;,a,-, • • • a¿ . It follows that V. , A.= [íz¿1) V \ai2) V ... V [ain) Ç A,-j V A¿2 V • • • V Ain. Consequently ? C F' (A . V A . V • • • V A . ) = F' (A . ) U F' (A . ) u • • • U F' (A . ). — 7 12« Z _ Z«12 Z • 7 -, Zzz Hence the result.
Theorem 6. The closure of the set of T. points of (ÍP, T) is F{D) where D is the dual ideal of dense elements of L.
Proof. By Lemma II, every maximal dual ideal of L is an element of j . It is easily seen that the maximal dual ideals are precisely the T points of (J\ T). Also, by Lemma III, D is the product of all the maximal dual ideals of L. Hence the result follows by Theorem 3. Theorem 7. (f, T) is Yl0 if D = [l).
Proof. Let g = F'{A) be a non-null open subset of (j , T) and let D = [l). Then A =¿ [l) and so there exists a maximal dual ideal M of L not containing A. (For, otherwise by Lemma III, A Ç D = [l).) Also, by Lemma II, it follows that M £ J. Thus g contains the closed set consisting of the single point M. Hence (5\ t) is n0.
Theorem 8. (j , T) is normal if and only if L is an S-lattice.
Proof. Suppose L is an S-lattice. Let Fl = F {A), F2 = F (B) be any two disjoint closed subsets of J. Then, by Theorem 1, F{A V B) =
= F([a) V [fl*)) = F([flfl*)) = F([0)) = F{L) = cp;
= F{B) n F([a* •)) C F([a*)) H F([fl* »))
= F(U') V [fl**)) = F([a*fl**)) = F([0)) = F(L) = r/..
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Since a** > a, a** £ A and so F}¡2Fv As a* £ B, F4 2 F2' Also F3 u F4 = F([a**))U F([a*)) = F([a**)n la*)) = F([«** + a*)) = F([l)) « ?. From the above it follows that (9, T) is normal. Conversely, suppose (9, T) is normal and x £ L. Set Fx = F ([x)) and F2 = F([x*)). Then clearly Fx n F2 = cf>. Hence, as (9, T) is normal, there exist closed subsets F? = F (A), F4 = F(B) containing Fx, F2 respectively such that (1) FjnF4 = 0, (2) F2nFi =
From (1) we have [x) V B = L. Hence xb = 0 for some b £ B. Clearly (4) x* £ fc.
Similarly, using (2), we can prove that, for some a £ A, (5) x** > a.
From (3) we have An B = [l). Hence [a) n [&) = [l) and so a + b = 1. From (4) and (5) it follows that x* + x** = 1. Thus L is an 5-lattice. Let us denote the set of all maximal dual ideals and the set of all dense max- imal dual ideals of L by JR and JR. respectively. By Lemma II, JR Q 9. Clearly every point of JR is Tx. Hence we have the following
Theorem 9. The subspaces (JR, T) and (JR., T) are Ty
Theorem 10. The subspace (JR, T) is the smallest of the subspaces X of (f, T) such that X is not weakly separable from any point outside it.
Proof. Let A £ 9 and A i%. By Lemma I, there exists M £ JR such that M 2 A. Thus MeSflF(A)andso)In F(A) ji
Theorem 11. !)His closed in (9, T). Proof. By Theorem 3, clDR= F(D). Let A £ dull. We shall prove that A £ JH. Let x £ L and x / A. Since (x + x*)* = 0, x + x*eDCA. Hence, as A is prime and x / A, x* £ A. Consequently, 0 = xx*e A V [x) and so A V [x) = L. Hence Ael. It follows that cl3H = JR.. Hence the result.
Corollary. The subspace (JR, T) is compact.
Theorem 12. If X is any subset of 9 containing JR, then (X, T) is compact.
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Proof. Let X C Uie¡F'{Ai). Then X C F' { V¿e/A¿) and so no member of X contains V^j/L. In particular no member of %. contains V¿E/^¿. By Lemma I, it follows that V¿e/A¿=L. Hence 0 a «¿¡Äi. • • • «l (for some fl¿eA¿.) so that L = [fl¿2fl¿2 • • • fl¿n) = [fl¿1) V [a,-2) V • • • V taif!) C A¿1 VA¿2V- • -VA¿n. It follows that
XCF'(A¿1 V Ai2 V...VAin) = F'(Ail) U F' (A¿2) u • • • M F*MiJ,
Hence the result.
Theorem 13. The subspace Oft.,, T) is compact.
Proof. Let Dj be the product of all the members of 5lîj. Then the closure of ÎHj in (511,T) is clearly ÏRnFiDj. Let M e 51Î- 3lTj. Suppose AODj. As Me Jlï, zMç2* M. for all M. £M^ and so, as the M. are prime, M* C M. for every M. e üllj (M* denotes the pseudocomplement of M in the lattice of dual semi-ideals of L). Hence M* C D, C M by assumption. Consequently, M* = [l) and so M has a pseudocomplement M in J) and M = [l). Thus M is dense. This is a contra- diction to our assumption that M fÈ ÜHj. Hence iM| Dj. It follows that every mem- ber of M which contains Dj is dense and so m n F(Dj) = 51L. Thus Jllj is closed in 0R, T). As (51Î, F) is compact it follows that 01Î,, T) is also compact. For any ideal A of L, let G(A) denote the set of all prime dual ideals dis- joint with A. Let G' (A) = J - G (A). Then we have the following
Theorem 14. (i) G(Vze/A¿)= C\i€lG{At), (ii) G(A 1 n A2C\ ■■ . n Aj = G(Aj) uG(A2)u---UG(An), (iii) G(L) = 0, (iv) G((0])= 3\ (Here the A . are ideals of L.)
Proof, (i) It is easily seen that for any prime dual ideal B of L, B O ( \Ji j A.) = (p'^'B D A. = cp for every z £ /. Hence (i). (ii) It is enough to prove that for a prime dual ideal B, BnAlC\A2n---OA =
Br\A1r\A2r\...nAn = (p for every i (= 1, 2,. • ■, n). Suppose B C\ A . ^
B n A. = cp for some z (= 1, 2, • • • , n). The proof of (ii) is now complete. (iii) and (iv) are obvious. The above theorem shows that G defines a closure operation in J , thereby giving rise to a topology T on J . Since G' (A) = f - G(A), as a consequence of Theorem 14 we have the fol- lowing
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Theorem 15. (i) C ( V^,^) = Ufe/ C (A), (ii) G' (Ax n A2 D ... n AJ = C (Aj) n g' (A2) n •.. n g' (ab), (iii) g' U) = 9, (iv) c' ((o]) = <¿.
From Theorems 14 and 15 we have the following
Theorem 16. The lattice of all open (closed) subsets of (9, T') is homomor- phic (dually homomorphic) to the lattice of ideals of L and the mapping A—*G'(A) (A —» G (A)) takes arbitrary lattice sums into the corresponding set unions (set in- tersections).
Theorem 17. // X is a subset of (9, T'), then clX = G(XQ), where XQ is the product of the set complements of all the members of X.
The proof of Theorem 17 is similar to the corresponding result of [15].
Theorem 18. The T. points of (9, T) are identical in their totality with the anti-T. points of (j, T1) and vice versa.
Proof. Suppose A is a Tx point of (9, T). Then the closure of A in (9, T) does not contain any other point of (S, T). Hence no other member of J contains A. In other words, A is not contained in any other member of J. Therefore A does not belong to the closure of any other point of (f, T'). Thus A is an anti- Tj point of (9, T1). The second part is proved on similar lines.
3. Bicomplemented lattices. This section is devoted to a study of the addi- tional features of the space (j , T) and its subspaces when the given lattice is bicomplemented. The results of this section extend the results of Balachandran [3] on Stone's topology of the distributive lattice to bicomplemented lattices. Throughout this section, L denotes a bicomplemented lattice. Since L is closed for quasicomplements, the lattice D of dual ideals of L is closed for pseudocomplements. Hence we can speak of simple dual ideals of L. The set of all simple maximal dual ideals of L is denoted by JR2<
Lemma 1. In a quasicomplemented lattice L, the pseudocomplement of a dual ideal A is the product of all the prime dual ideals not containing A.
Proof. Let An denote the pseudocomplement of A and B the product of